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A ray of light strikes a flat block of glass \((n=1.50)\) of thickness \(2.00 \mathrm{cm}\) at an angle of \(30.0^{\circ}\) with the normal. Trace the light beam through the glass, and find the angles of incidence and refraction at each surface.

Short Answer

Expert verified
The angle of refraction after entering the glass is given by \( r_1 = \arcsin[\frac{(n_i/n_r) \cdot \sin(i)} ] \). The light travels through the glass, hitting the internal surface at this same angle. After exiting the block, the angle becomes \( i_2 = \arcsin[\frac{(n_r/n_i) \cdot \sin(r_1)} ] \), assuming air as the outside medium with refractive index \( n_r = 1.00 \).

Step by step solution

01

Find Initial Refraction Angle

When the ray of light hits the top surface of the glass block, it refracts into the block. To find the angle of refraction \( r_1 \), we apply Snell's law, which states \( n_i \cdot \sin(i) = n_r \cdot \sin(r) \). Let's solve for \( r_1 \): Here, the incident angle \( i = 30.0^\circ \), the refractive index of air \( n_i = 1.00 \), and the refractive index of glass \( n_r = 1.50 \). By substituting these values in Snell's law and solving for \( r_1 \), we get \( r_1 = \arcsin[\frac{(n_i/n_r) \cdot \sin(i)} ]\).
02

Find Internal Reflection Angle

As the light beam travels through the glass, it hits the bottom surface at the same angle at which it entered, \( r_1 \), because the horizontal surfaces are parallel. This angle becomes the angle of incidence for the bottom surface.
03

Find Exit Refraction Angle

At the bottom surface, the ray refracts and exits the block. To find the exit angle, we apply Snell's law again. The exit angle \( i_2 \) will be equal to \( i_2 = \arcsin[\frac{n_r/n_i) \cdot \sin(r_1)} \] \). Here, \( n_i = 1.50 \) is now the refractive index of glass (since the light is inside the glass), and \( n_r = 1.00 \) is the refractive index of air (since the light is exiting into air).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refraction of Light
Refraction of light is a phenomenon that occurs when light travels from one transparent medium to another with a different density. This change in density alters the speed at which light travels, causing the light ray to change direction. The amount of bending or refraction that occurs is guided by Snell's Law. It is an everyday occurrence, noticeable when a straw appears bent in a glass of water or when a pool seems shallower than it actually is.

In our example, a ray of light encounters a dense medium, a glass block, from a less dense medium, air. Consequently, the path of the light ray bends towards the normal—the imaginary line perpendicular to the surface where light enters the glass. The degree of refraction not only depends on the angle at which light strikes the surface but also on the inherent properties of the materials involved, primarily their refractive indices, a concept we will explore shortly.
Angle of Incidence
The angle of incidence is the angle between the incident ray of light and the normal to the surface at the point of contact. The normal is an essential reference point as it allows us to measure the degree to which light bends when transitioning between different mediums. In our textbook problem, the light ray strikes the glass at a 30-degree angle from the normal—a moderate angle that ensures the light will bend but not reflect off the surface.

Understanding the angle of incidence is crucial for predicting the behavior of light. This angle plays a pivotal role when applying Snell's law, as it helps determine the angle of refraction. The angle of incidence remains constant when light passes through a substance with parallel surfaces—like our glass block—unless the medium itself changes.
Refractive Index
The refractive index, typically symbolized as 'n', indicates how much a substance can bend light. It's a ratio that compares the speed of light in a vacuum to its speed in the given material. For example, the refractive index of air is approximately 1.00 because light travels through air almost as fast as it does in a vacuum. In contrast, the refractive index of glass in our problem is 1.50, signaling that light travels more slowly in glass compared to air.The refractive index is a dimensionless number that plays a central role in Snell's law, a fundamental equation in optics. By knowing the refractive indices of both the original and the new mediums, we can predict exactly how much light will bend when entering or exiting a material. This property allows us to design lenses, glasses, and a multitude of optical instruments that improve and manipulate our vision and perception of the world around us.

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Most popular questions from this chapter

A narrow beam of light is incident from air onto the surface of glass with index of refraction \(1.56 .\) Find the angle of incidence for which the corresponding angle of refraction is half the angle of incidence. (Suggestion: You might want to use the trigonometric identity \(\sin 2 \theta=2 \sin \theta \cos \theta .)\)

Refer to Quick Quiz \(35.4 .\) By how much does the duration of an optical day exceed that of a geometric day? Model the Earth's atmosphere as uniform, with index of refraction \(1.000293,\) a sharply defined upper surface, and depth \(8614 \mathrm{m} .\) Assume that the observer is at the Earth's equator, so that the apparent path of the rising and setting Sun is perpendicular to the horizon.

A light ray initially in water enters a transparent substance at an angle of incidence of \(37.0^{\circ},\) and the transmitted ray is refracted at an angle of \(25.0^{\circ} .\) Calculate the speed of light in the transparent substance.

A prism that has an apex angle of \(50.0^{\circ}\) is made of cubic zirconia, with \(n=2.20 .\) What is its angle of minimum deviation?

Consider a common mirage formed by super-heated air just above a roadway. A truck driver whose eyes are \(2.00 \mathrm{m}\) above the road, where \(n=1.0003,\) looks forward. She perceives the illusion of a patch of water ahead on the road, where her line of sight makes an angle of \(1.20^{\circ}\) below the horizontal. Find the index of refraction of the air just above the road surface. (Suggestion: Treat this as a problem in total internal reflection.)

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